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The function $f(x) = x^x$

Let us look at this function 

$$f(x) = x^x \tag{1}$$

and try to find its derivative for real values of $x$. 

Before this we should mention that this function is well defined only when $x \gt 0$. 

In other words... to avoid any complications we are looking at this function only for positive values of $x$. 

I. How do we go about finding the derivative $f'(x) = (x^x)'$ ? 

Well, we will use the following identity 

$$a^x = e^{x \cdot \ln a} \tag{2}$$

which is well-known from high school math and which holds true for any $a \gt 0$ and any $x \in \mathbb{R}$. 

Substituting $a = x$ in this identity we subsequently get 

$$f(x) = x^x = e ^ {x \cdot \ln x}$$ 

$$f'(x) = e ^ {x \cdot \ln x} \cdot (x \cdot \ln x)'$$ 

$$f'(x) = e ^ {x \cdot \ln x} \cdot (1 \cdot \ln x + x \cdot \frac{1}{x})$$ 

$$f'(x) = e ^ {x \cdot \ln x} \cdot (\ln x + 1)$$ 

$$f'(x) = x ^ x \cdot (\ln x + 1) \tag{3}$$ 

The last equation $(3)$ gives us the derivative which we wanted to find. 

In the above derivation we used several simple facts from math analysis. 

$$(e^x)' = e^x$$

$$(f(g(x)))' = f'(g(x)) \cdot g'(x)$$

$$(u(x) \cdot v(x))' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$$ 

Finally let us restate the above established formula.   

$$ \large { (x^x)' = x ^ x \cdot (\ln x + 1) } \tag{4}$$ 

II. How do we find $\displaystyle{ \lim_{x \to 0^{+}} f(x) } = \displaystyle{ \lim_{x \to 0^{+}} x^x }$ ?

I think the easiest way of finding this limit which I have seen is by letting $x=e^{-t}$ where $t$ is some very large positive number. Then we easily get the following. 

$$\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} x^x = \lim_{x \to 0^{+}} e ^ {x \cdot \ln x} = \lim_{x \to 0^{+}} e ^ {(\ln x) \cdot x} =  $$

$$ = \lim_{t \to \infty} e^{(-t) \cdot e^{-t} } = \lim_{t \to \infty} e^{ \frac{(-t)}{e^{t}} } = e ^ {\lim_{t \to \infty} \frac{(-t)}{e^{t} } } = e^0 = 1 $$

Here the main fact which we used was that  

$$\lim_{t \to \infty} \frac{(-t)}{e^{t} }  = 0$$ 

which is quite obvious given that the numerator is a polynomial of $t$ and the denominator is the exponential function $e^t$.  

This way we have just calculated this quite remarkable limit 

$$ \large \lim_{x \to 0^{+}} x^x = 1 \tag{5}$$ 

Finally... here is a video demonstrating nicely an informal numerical approach to finding the same limit What is 0 to the power of 0?

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